Evaluate the following integral involving $\sin \pi x$

Let's write the function $F(\sin\pi x)$ like this $$F(\sin\pi x)= \begin{cases}1, & \text{if }\sin\pi x\ge \dfrac12 \\[0.2cm]0, & \text{if }\sin\pi x< \dfrac12 \end{cases}$$ Now we solve the inequality $\sin\pi x \ge \frac12$. We only care about the solutions in the range $[0;1]$. You already did it. It holds for $x \in [\frac16;\frac56]$. So we can rewrite the function $F(\sin\pi x)$ like this: $$F(\sin\pi x)= \begin{cases}1, & \text{if } x \in \big[\frac16;\frac56\big] \\[0.2cm]0, & \text{if } x \in \big[0;\frac16\big)\cup\big(\frac56;1\big] \end{cases}$$ Now we can split the integral to two parts, one for each case of this function: $$ \int_0^1 F(\sin\pi x)dx = \int_{[0;1]} F(\sin\pi x)dx =\\= \int_{\big[\frac16;\frac56\big]}F(\sin\pi x)dx + \int_{\big[0;\frac16\big)\cup\big(\frac56;1\big]}F(\sin\pi x)dx = \int_{\big[\frac16;\frac56\big]}1dx + \int_{\big[0;\frac16\big)\cup\big(\frac56;1\big]}0dx $$ If you aren't familiar with this notation, it is the same as: $$ \int_\frac16^\frac561dx+\left(\int_0^\frac160dx+\int_\frac56^10dx\right) $$ The second term is zero, so only the first term remains. And you see that your result was correct.